/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 A space probe \(2.0 \times 10^{1... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A space probe \(2.0 \times 10^{10} \mathrm{~m}\) from a star measures the total intensity of electromagnetic radiation from the star to be \(5.0 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\). If the star radiates uniformly in all directions, what is its total average power output?

Short Answer

Expert verified
To calculate the total average power output of the star, you first calculate the spherical surface area at the location of the probe. Using this area and the given intensity, you find the total power output by multiplying the intensity and the surface area.

Step by step solution

01

Understand the problem and what is given

The intensity \(I\) of the electromagnetic radiation is \(5.0 \times 10^{3} \mathrm{~W/m^{2}}\) where the distance \(r\) between the star and the probe is \(2.0 \times 10^{10} \mathrm{~m}\). The problem involves calculating the total power \(P\) output of the star.
02

Apply the equation for the surface area of a sphere

The surface area \(A\) of a sphere can be described as \(A = 4\pi r^2\). Let's put the value of \(r\) into the equation: \(A = 4\pi (2.0 \times 10^{10} \mathrm{~m})^2\).
03

Calculate the total power output

Power \(P\) can be described as the product of intensity \(I\) and the area \(A\) over which it is spread. Thus the total power output of the star will be \(P = IA\). Substituting \(I\) and \(A\) into the equation will give us the total power output.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Output
Power output is a critical concept in physics, especially when discussing electromagnetic radiation from stars and other celestial bodies. When we mention power output, we mean the total amount of energy a star emits per unit time. This is usually measured in watts (W). It represents how much energy is distributed into the surrounding space and affects various factors, such as the light intensity reaching planets and probes.

Consider a star radiating energy uniformly in all directions. The power output can help us understand the vastness of its energy emission. The more powerful a star, the higher its ability to illuminate and heat surrounding areas. This understanding is key when studying how stars influence their nearby planets and their ability to support life.

In our context, power output is derived from knowing the intensity (energy per unit area) and applying it to the geometry of a sphere, which will be our next concept. This transformation helps simplify many astronomical calculations.
Intensity Calculation
Intensity is a measure of how much power is received per unit area, measured in \( ext{W/m}^{2} \). When dealing with electromagnetic radiation, like light from a star, intensity tells us how dense or concentrated this energy is over a given surface area. This density of energy is crucial in understanding how bright or warm an object receiving this radiation will feel.

Calculating intensity involves knowing how much power is available and over what area it spreads. As stars emit energy in all directions, understanding this spread becomes essential. In our problem, we already know the measured intensity at a known distance from the star. This information allows us to calculate how much total power the star is emitting.

By working backwards from the measured intensity at a specific location (like our space probe position), we can determine the total power output of the star, which turns out to be a fundamental aspect of astronomy.
Surface Area of a Sphere
The surface area of a sphere plays a vital role when dealing with celestial objects radiating energy evenly in all directions, such as stars. Mathematically, the surface area \( A \) is given by the formula \( A = 4 \pi r^{2} \), where \( r \) is the radius of the sphere.

In this context, the sphere represents an imaginary shell at a fixed distance from the star, specifically the distance to the space probe. This formula becomes handy as it helps calculate how the star's emitted energy is distributed over space.

When you know this surface area, you can use it alongside intensity to find the total power output, as outlined in the exercise. By multiplying intensity by the surface area, you can derive the total energy the star emits, giving a clear insight into its energy properties. Understanding this calculation provides a stronger grasp of how energy disperses in space, impacting planets and other objects around the star.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm{~km} ?\) (b) Light from the star Sirius takes 8.61 years to reach the earth. What is the distance from earth to Sirius in kilometers?

A monochromatic light source with power output \(60.0 \mathrm{~W}\) radiates light of wavelength \(700 \mathrm{nm}\) uniformly in all directions. Calculate \(E_{\max }\) and \(B_{\max }\) for the \(700 \mathrm{nm}\) light at a distance of \(5.00 \mathrm{~m}\) from the source.

An electromagnetic wave with frequency \(65.0 \mathrm{~Hz}\) travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. The electric field has amplitude \(7.20 \times 10^{-3} \mathrm{~V} / \mathrm{m} .\) (a) What is the speed of propagation of the wave? (b) What is the wavelength of the wave? (c) What is the amplitude of the magnetic field?

An air-filled cavity for producing electromagnetic standing waves has two parallel, highly conducting walls separated by a distance \(L\). One standing- wave pattern in the cavity produces nodal planes of the electric field with a spacing of \(1.50 \mathrm{~cm}\). The next-higher-frequency standing wave in the cavity produces nodal planes with a spacing of \(1.25 \mathrm{~cm} .\) What is the distance \(L\) between the walls of the cavity?

The energy flow to the earth from sunlight is about \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). (a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity. (b) The distance from the earth to the sun is about \(1.5 \times 10^{11} \mathrm{~m} .\) Find the total power radiated by the sun.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.